The nuclei below are observed to decay by emitting an particle. Write out the de

Question

The nuclei below are observed to decay by emitting an particle. Write out the decay process for each of these nuclei, and determine the energy released in each reaction. Be sure to take into account the mass of the electrons associated with the neutral atoms. (Type the decay processes using the format 14C6 for 14 6 C and any symbols from the following table.) summary of nuclear particles; consult your textbook for relevant examples (a) 243 95 Am 239 Pu 93+ Incorrect: Your answer is incorrect. E = MeV (b) 232 90 Th $$226Ra88+a Incorrect: Your answer is incorrect. E = MeV

Explanation / Answer

(a)

The given decay process is not correct. After alpha decay, the atomic number of Am should decreased by 2 units. the product shoud be Np nucleus and He nucleus. Thus, the correct decay process is

243 Am 95 239 Np 93 +

Now, the alpha particle is a helium nucleus. Thus, the decay process can again be represented as

243 Am 95 239 Np 93 + 4 He 2

The atomic mass of the 239 Am 95 is

MAm = 243 u

Here, atomic mass unit is u.

The atomic mass of the 239 Np 93 is

MNp = 239 u

The atomic mass of 2He4 is

MHe = 4.0026 u

Now,

Sum of the masses of reactants is

MAm = 239 u

Sum of the masses of products is

MHe + MNp = 239 u + 4.0026 u

Now,

M = Sum of the mass of reactant – Sum of the masses of products

       = 243 u – (239 u + 4.0026 u)

       = – 0.0026 u

Negative indicates that the decay process is endoergic.

Thus, the energy released in the reaction is

E = (0.0026 u) (931.5 MeV / 1 u)

   = 2.4219 MeV

Rounding off to three significant figures, the energy released in the decay process is 2.42 MeV.

(b)

The given decay process is not correct. After alpha decay, the atomic number of Th should decreased by 2 units. Thus, the correct decay process is

232 Th 90 228 Ra 88 +

Now, the alpha particle is a helium nucleus. Thus, the decay process can again be represented as

232 Th 90 228 Ra 88 + 4 He 2

The atomic mass of the 232 Th 90 is

MTh = 232.0381 u

Here, atomic mass unit is u.

The atomic mass of the 228 Ra 88 is

MRa = 228 u

The atomic mass of 2He4 is

MHe = 4.0026 u

Now,

Sum of the masses of reactants is

MTh = 232.0381 u

Sum of the masses of products is

MRa + MHe = 228 u + 4.0026 u

Now,

M = Sum of the mass of reactant – Sum of the masses of products

       = 232.0381 u – (228 u + 4.0026 u)        

       = 0.0355 u

Thus, the energy released in the reaction is

E = (0.0355 u) (931.5 MeV / 1 u)

   = 33.06825 MeV

Rounding off to three significant figures, the energy released in the decay process is 33.1 MeV.

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